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Mar 12, 2014 at 19:50 vote accept Ludolila
Mar 10, 2014 at 20:11 comment added user46855 @Ludolila sorry, my poor English made me understand "(compact) or (locally compact and locally connected)"
Mar 10, 2014 at 8:19 comment added Ludolila @user46855 I took $X$ to be locally compact and locally connected (and Hausdorff). In this case the compact open topology turns $H(X)$ into a topological group. See for example "Topologies for homeomorphisms groups" by Arens. I will check Comfort, though, sounds interesting.
Mar 9, 2014 at 22:49 comment added user46855 According to Comfort (chap. 24 in handbook of set-theoretic topology, pag. 1242) citing de Groot, the groups of homeomorphism of compact connected spaces are arbitrary groups, and so the discrete topology might be the only group topology. @Ludolila might want stronger conditions on $X$ to have a topological group $H(X)$
Mar 9, 2014 at 20:24 history edited Ludolila CC BY-SA 3.0
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Mar 9, 2014 at 20:23 comment added Ludolila @JohnPardon: Just to make sure: if $C(X)$ is a topological vector space, then it will be path connected. So the answer to the question "which topologies make $C(X)$ connected" is "the topologies that make $C(X)$ a topological vector space"?
Mar 9, 2014 at 19:57 comment added Ludolila @JosephVanName No, I haven't even heard about it. Just finished reading about it in Wikipedia, and it's very interesting! Thanks for the idea!
Mar 9, 2014 at 19:45 comment added John Pardon @Ludolila: well then $C(X)$ is a vector space and thus contractible.
Mar 9, 2014 at 19:43 comment added Ludolila @JohnPardon $C(X)$-the space of all continuous functions on $X$.
Mar 9, 2014 at 19:12 comment added Joseph Van Name @Ludolila. Have you read about the mapping class group? Essentially the mapping class group of a space is the group of autohomeomorphisms modulo the subgroup of autohomeomorphisms isotopic to the identity. In particular, the mapping class group is the collection of all path components in $Aut(X)$.
Mar 9, 2014 at 19:08 answer added Ali Taghavi timeline score: 1
Mar 9, 2014 at 19:03 comment added John Pardon The specific question about connectedness of $H_+(S^1)$ is easy to treat explicitly: you can just write down a path between any homeo and the identity.
Mar 9, 2014 at 19:02 comment added John Pardon What is $C(X)$?
Mar 9, 2014 at 18:46 history asked Ludolila CC BY-SA 3.0